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      leetcode733图像渲染刷题笔记
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        <h2 id="图像渲染"><a href="#图像渲染" class="headerlink" title="图像渲染"></a><a target="_blank" rel="noopener" href="https://leetcode.cn/problems/flood-fill/">图像渲染</a></h2><blockquote>
<p>有一幅以 m x n 的二维整数数组表示的图画 image ，其中 image[i][j] 表示该图画的像素值大小。<br>你也被给予三个整数 sr ,  sc 和 newColor 。你应该从像素 image[sr][sc] 开始对图像进行 上色填充 。<br>为了完成 上色工作 ，从初始像素开始，记录初始坐标的 上下左右四个方向上 像素值与初始坐标相同的相连像素点，接着再记录这四个方向上符合条件的像素点与他们对应 四个方向上 像素值与初始坐标相同的相连像素点，……，重复该过程。将所有有记录的像素点的颜色值改为 newColor 。<br>最后返回 经过上色渲染后的图像 。</p>
</blockquote>
<h3 id="题目分析"><a href="#题目分析" class="headerlink" title="题目分析"></a>题目分析</h3><p>列出自己想到的线索</p>
<ul>
<li>维护一个存储要检查的点的集合，toBeChecked</li>
</ul>
<h3 id="根据题解进行分析"><a href="#根据题解进行分析" class="headerlink" title="根据题解进行分析"></a>根据题解进行分析</h3><ul>
<li>题解里看到这道题的最佳解法是使用DFS(深度优先搜索)</li>
<li>搜索了一下DFS，DFS里可以通过自己调用自己的方法，也就是递归来实现</li>
</ul>
<h3 id="递归"><a href="#递归" class="headerlink" title="递归"></a>递归</h3><ul>
<li>递归的特点就是自己调用自己</li>
<li>递归由两部分组成：（1）递归操作 （2）递归终止条件</li>
</ul>
<h4 id="一个最简单的递归"><a href="#一个最简单的递归" class="headerlink" title="一个最简单的递归"></a>一个最简单的递归</h4><ul>
<li>计算从1，2，3公差为1的等差数列前n项和的程序<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * 递归操作，进行一次加法操作</span></span><br><span class="line"><span class="comment"> * 递归终止条件，当前的数字等于1</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="keyword">public</span> <span class="type">int</span> <span class="title function_">sum</span><span class="params">(<span class="type">int</span> n)</span> &#123;</span><br><span class="line">    <span class="type">int</span> <span class="variable">sum</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">if</span> (n == <span class="number">1</span>) &#123;</span><br><span class="line">        sum += <span class="number">1</span>;</span><br><span class="line">    &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">        sum = sum(n - <span class="number">1</span>) + n;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> sum;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></li>
</ul>
<h4 id="递归增加"><a href="#递归增加" class="headerlink" title="递归增加"></a>递归增加</h4><ul>
<li>计算从1，7，13公差为6的等差数列前n项和的程序<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="type">int</span> <span class="title function_">sum6</span><span class="params">(<span class="type">int</span> n)</span>&#123;</span><br><span class="line">    <span class="type">int</span> <span class="variable">sum</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">if</span>(n == <span class="number">1</span>)&#123;</span><br><span class="line">        sum += n;</span><br><span class="line">    &#125;<span class="keyword">else</span>&#123;</span><br><span class="line">        sum = sum6(n - <span class="number">6</span>) + n;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> sum;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></li>
</ul>
<h4 id="使用递归来解决这道题"><a href="#使用递归来解决这道题" class="headerlink" title="使用递归来解决这道题"></a>使用递归来解决这道题</h4><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@param</span> image 二维数组来代表原始的图片</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@param</span> sr 目标点的行号</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@param</span> sc 目标点的列号</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@param</span> color 目标点的新颜色</span></span><br><span class="line"><span class="comment"> * <span class="doctag">@return</span> 上好色的图片</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="keyword">public</span> <span class="type">int</span>[][] floodFill(<span class="type">int</span>[][] image, <span class="type">int</span> sr, <span class="type">int</span> sc, <span class="type">int</span> color) &#123;</span><br><span class="line">    <span class="type">int</span> <span class="variable">oldColor</span> <span class="operator">=</span> image[sr][sc];</span><br><span class="line">    dfs(image, oldColor, sr, sc, color);</span><br><span class="line">    <span class="keyword">return</span> image;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * 递归</span></span><br><span class="line"><span class="comment"> * 递归操作：</span></span><br><span class="line"><span class="comment"> * 1. dfs 行号 - 1</span></span><br><span class="line"><span class="comment"> * 2. dfs 行号 + 1</span></span><br><span class="line"><span class="comment"> * 3. dfs 列号 - 1</span></span><br><span class="line"><span class="comment"> * 4. dfs 列号 + 1</span></span><br><span class="line"><span class="comment"> * 递归终止条件：</span></span><br><span class="line"><span class="comment"> * 1. 当前的行号&lt;0</span></span><br><span class="line"><span class="comment"> * 2. 当前的列号&lt;0</span></span><br><span class="line"><span class="comment"> * 3. 当前的行号&gt;行长</span></span><br><span class="line"><span class="comment"> * 4. 当前的列号&gt;行宽</span></span><br><span class="line"><span class="comment"> * 5. 当前的颜色不等于oldColor</span></span><br><span class="line"><span class="comment"> * 6. 当前的颜色等于新颜色</span></span><br><span class="line"><span class="comment"> * return;</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="keyword">public</span> <span class="keyword">void</span> <span class="title function_">dfs</span><span class="params">(<span class="type">int</span>[][] image, <span class="type">int</span> oldColor, <span class="type">int</span> sr, <span class="type">int</span> sc, <span class="type">int</span> newColor)</span> &#123;</span><br><span class="line">    <span class="type">int</span> <span class="variable">rLength</span> <span class="operator">=</span> image.length;</span><br><span class="line">    <span class="type">int</span> <span class="variable">cLength</span> <span class="operator">=</span> image[<span class="number">0</span>].length;</span><br><span class="line">    <span class="keyword">if</span>(sr &lt; <span class="number">0</span> || sc &lt; <span class="number">0</span> || sr &gt;= rLength || sc &gt;= cLength || image[sr][sc] != oldColor || image[sr][sc] == newColor)&#123;</span><br><span class="line">        <span class="keyword">return</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    image[sr][sc] = newColor;</span><br><span class="line"></span><br><span class="line">    dfs(image, oldColor, sr - <span class="number">1</span>, sc, newColor);</span><br><span class="line">    dfs(image, oldColor, sr + <span class="number">1</span>, sc, newColor);</span><br><span class="line">    dfs(image, oldColor, sr, sc - <span class="number">1</span>, newColor);</span><br><span class="line">    dfs(image, oldColor, sr, sc + <span class="number">1</span>, newColor);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<h3 id="二叉树的前序遍历、中序遍历、后续遍历"><a href="#二叉树的前序遍历、中序遍历、后续遍历" class="headerlink" title="二叉树的前序遍历、中序遍历、后续遍历"></a>二叉树的前序遍历、中序遍历、后续遍历</h3><ul>
<li>dfs深度优先搜索最基础的应用应该就是用在二叉树的前中后序遍历上</li>
<li>如下为通过递归实现二叉树前序遍历的代码<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * Definition for a binary tree node.</span></span><br><span class="line"><span class="comment"> * public class TreeNode &#123;</span></span><br><span class="line"><span class="comment"> *     int val;</span></span><br><span class="line"><span class="comment"> *     TreeNode left;</span></span><br><span class="line"><span class="comment"> *     TreeNode right;</span></span><br><span class="line"><span class="comment"> *     TreeNode() &#123;&#125;</span></span><br><span class="line"><span class="comment"> *     TreeNode(int val) &#123; this.val = val; &#125;</span></span><br><span class="line"><span class="comment"> *     TreeNode(int val, TreeNode left, TreeNode right) &#123;</span></span><br><span class="line"><span class="comment"> *         this.val = val;</span></span><br><span class="line"><span class="comment"> *         this.left = left;</span></span><br><span class="line"><span class="comment"> *         this.right = right;</span></span><br><span class="line"><span class="comment"> *     &#125;</span></span><br><span class="line"><span class="comment"> * &#125;</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line">    <span class="keyword">private</span> List&lt;Integer&gt; list = <span class="keyword">new</span> <span class="title class_">ArrayList</span>&lt;&gt;();</span><br><span class="line"></span><br><span class="line">    <span class="keyword">public</span> List&lt;Integer&gt; <span class="title function_">preorderTraversal</span><span class="params">(TreeNode root)</span> &#123;</span><br><span class="line">        <span class="keyword">if</span>(root != <span class="literal">null</span>)&#123;</span><br><span class="line">            list.add(root.val);</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">if</span> (root != <span class="literal">null</span> &amp;&amp; root.left != <span class="literal">null</span>) &#123;</span><br><span class="line">            preorderTraversal(root.left);</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">if</span> (root != <span class="literal">null</span> &amp;&amp; root.right != <span class="literal">null</span>) &#123;</span><br><span class="line">            preorderTraversal(root.right);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> list;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></li>
</ul>
<h3 id="八皇后问题"><a href="#八皇后问题" class="headerlink" title="八皇后问题"></a>八皇后问题</h3><ul>
<li>题目<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line">    <span class="keyword">public</span> List&lt;List&lt;String&gt;&gt; <span class="title function_">solveNQueens</span><span class="params">(<span class="type">int</span> n)</span> &#123;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></li>
</ul>
<h4 id="题意"><a href="#题意" class="headerlink" title="题意"></a>题意</h4><ul>
<li>入参n代表棋盘行的个数，列的个数，要摆放的皇后的个数</li>
<li>输出字符串集合，点表示不放置棋子，Q表示放置皇后</li>
</ul>
<h4 id="解题思路"><a href="#解题思路" class="headerlink" title="解题思路"></a>解题思路</h4><ul>
<li>定义属性，二维数组board[n][n], 解法编号count</li>
<li>定义行为，isValid(), printBoard</li>
</ul>

      
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